Tipped and Partially Filled Frustum

Date: 12/14/2003 at 00:37:23
From: Matty
Subject: Find volume of portion of Frustum of a Cone
G'day doc,
I work in a processing plant and I need to find the volume of a
vessel. One section of the vessel is the frustum of a cone on its
side. A liquid is contained in this section and pours out the end of
the cone section, therefore the liquid only takes up a certain portion
of the cone's volume. The level of the liquid remains constant within
the cone. Both base and top radius are known, as is the height of the
cone section and the height of the liquid above the edge of both base
and top circles.
I have found the formula for the frustum of a cone and also the for a
section of a circle. But I'm not sure how to combine the two.

Date: 12/15/2003 at 06:28:53
From: Doctor Jeremiah
Subject: Re: Find volume of portion of Frustum of a Cone
Hi Matty,
Let's assume that the liquid level is constant at something over half
full and not quite as high as the top of the smaller circle. Then
from the side the tank, it would look like this:
__
+ \
+ \
_ + air \
/ \ \
|---|-----------| ---
| | | d
----+ | |--- --- --- ---
| | liquid | | |
| | | m |
\_/ / | |
+ / --- n
+ / |
+__/ |
---
|----h----|
There is a lot of mathematics in this answer. If you get lost you can
always just go straight to the answer at the bottom. But if you get
confused about anything and want to understand it better, then please
write me back and ask.
Many answers that calculate volume do so by adding up thin slices.
This approach is calculus. Calculus works well for this question
because each thin vertical slice is actually a section of a circle.
An arbitrary vertical slice would look like this:
+---------------+ ---
+ + |
+ + d
+ + |
+ + + ---
+ +
+ +
+ +
+ +
+++
If we label it slightly differently we get:
+-------+-------+
+ + | + +
+ R d R +
+ +a|a+ +
+ + + ---
+ + |
+ + |
+ + R
+ + |
+++ ---
where R is the radius of the circle and a is the angle of the triangle
made by the lines I drew.
The area of this shape is the sum of the area of the partial circular
part plus the area of the two triangles.
The partial circular part is the whole circle minus the fraction that
includes the triangles. If we knew what angle the two triangles took
then that fraction would be that value divided by 360 degrees.
We can find what angle the triangles take up with trigonometry:
+------L----+
+ |
+ d
R |
+ a|
+ |
+
In particular, we can use the definition of the cosine function:
cos(a) = d/R
If we solve that for the angle we get:
a = arccos(d/R)
The arc cosine is the inverse of the cosine. To calculate the cosine
and the arc cosine you need a calculator with trigonometric functions.
(Microsoft Windows comes with one, so if you use Microsoft Windows,
then you can select Calculator from the Accessories menu under
Programs on the Start menu. After starting the Calculator program, to
enable the trigonometric functions you need to select Scientific from
the View menu of the program.)
The angle of the two triangles together is 2a so the fraction of the
circle taken up by the part that includes the two triangles is:
2a/360 = arccos(d/R)/180
And this works perfectly when a is in degrees and arccos(d/R) returns
an answer in degrees. But the integral I do later relies on arccos
(d/R) returning a value in radians from 0 to 2Pi. So we need to
change this to assume that a is in radians and arccos(d/R) will be
returning a value in radians:
2a/(2Pi) = a/Pi = arccos(d/R)/Pi
And the area of the partial circular part is the area of the whole
circle minus the area of that fraction:
Pi R^2 - Pi R^2 * arccos(d/R)/Pi
Which is:
Pi R^2 - R^2 * arccos(d/R)
Now we need the area of the triangles:
+------L----+
+ |
+ d
R |
+ a|
+ |
+
The area of each triangle is:
Ld/2
Where L can be determined using the Pythagorean Theorem:
R squared = d squared + L squared
If we arrange this to find L we get:
L = square_root( R squared - d squared )
If I can use ^ to mean "with an exponent of" then ^2 means "with an
exponent of 2" or "squared" and if I use sqrt as a short form of
square_root then we can shorten that to look like this:
L = sqrt( R^2 - d^2 )
So the area of each triangle is:
d * sqrt( R^2 - d^2 ) / 2
And the area of both triangles is:
d * sqrt( R^2 - d^2 )
So the area of the whole shape is the sum of the area of the partial
circular part plus the area of the two triangles:
A = Pi R^2 - R^2 * arccos(d/R) + d * sqrt( R^2 - d^2 )
That is the equation for a area of a circle with a chord sliced off
(where arccos(d/R) returns an angle in radians). Now that we have
the area of a thin slice we can find the volume of whole thing by
adding up the volume of a whole bunch of thin slices. That's the way
calculus works. But we need the volume of a really thin slice. If
we say the thickness of this really thin slice is dx then the volume
of the slice is:
dV = A * dx
And the total volume is the sum of all the slices from the point where
R equals m to the point where R equals n. That is called an integral
and is drawn like this:
R=n
/
V = | dV
/
R=m
We can write that like this:
R=n
/
V = | A dx
/
R=m
And if we stick in the value of A we get:
R=n
/
V = | [Pi R^2 - R^2 arccos(d/R) + d sqrt(R^2-d^2)] dx
/
R=m
We can apply the integral to smaller pieces of the equation using the
distributive property:
R=n
/
V = | Pi R^2 dx
/
R=m
R=n
/
- | R^2 arccos(d/R) dx
/
R=m
R=n
/
+ | d sqrt(R^2-d^2) dx
/
R=m
The line that defines the outside of the cone is:
| + R = slope x
| +
| +
| +
| - - - - - - - + - - - +---h---+ - - - - -
| + | |
| + d d
| + b | |
---+----------------------------------
| + | |
| + | |
| + m n
| + | |
| + | |
| + |
| + |
| +
Where:
slope = tan(b) = (n-m)/h
The slope is (n-m)/h, but the slope is also a really small increase
in R (which we will call dR) divided by a really small increase in x
(which we will call dx) so:
dR/dx = (n-m)/h
And:
dx = dR h/(n-m)
So we can stick that into our integral to make:
R=n
/
V = | Pi R^2 dR h/(n-m)
/
R=m
R=n
/
- | R^2 arccos(d/R) dR h/(n-m)
/
R=m
R=n
/
+ | d sqrt(R^2-d^2) dR h/(n-m)
/
R=m
Which can be simplified to:
R=n
/
V = Pi * h/(n-m) | R^2 dR
/
R=m
R=n
/
- h/(n-m) | R^2 arccos(d/R) dR
/
R=m
R=n
/
+ d * h/(n-m) | sqrt(R^2-d^2) dR
/
R=m
Now, we could solve this integral ourselves (which is too much work)
or we could let computers do it for us. We can use a program at
http://integrals.wolfram.com/
to solve these integrals. The only thing is that the program wants
the equation in terms of x so when typing equations into that site
replace each R with x and each dR with dx. And then change the x in
the answer back to R.
This is the answer that site gives:
V(R) = Pi * h/(n-m) * R^3/3
+ h/(n-m) * Rd/6 * sqrt(R^2 - d^2)
- h/(n-m) * R^3/3 * arccos(d/R)
+ h/(n-m) * d^3/6 * log(R + sqrt(R^2 - d^2))
+ d * h/(n-m) * R/2 * sqrt(R^2 - d^2)
- d * h/(n-m) * d^2/2 * log(R + sqrt(R^2 - d^2))
where R varies from m to n. If we stick in those boundaries we get
the actual volume:
Volume = V(n) - V(m)
Which is:
V(n) = Pi * h/(n-m) * n^3/3
+ h/(n-m) * nd/6 * sqrt(n^2 - d^2)
- h/(n-m) * n^3/3 * arccos(d/n)
+ h/(n-m) * d^3/6 * log(n + sqrt(n^2 - d^2))
+ d * h/(n-m) * n/2 * sqrt(n^2 - d^2)
- d * h/(n-m) * d^2/2 * log(n + sqrt(n^2 - d^2))
V(m) = Pi * h/(n-m) * m^3/3
+ h/(n-m) * md/6 * sqrt(m^2 - d^2)
- h/(n-m) * m^3/3 * arccos(d/m)
+ h/(n-m) * d^3/6 * log(m + sqrt(m^2 - d^2))
+ d * h/(n-m) * m/2 * sqrt(m^2 - d^2)
- d * h/(n-m) * d^2/2 * log(m + sqrt(m^2 - d^2))
So the volume of the liquid in your tank is:
Volume = Pi * h/(n-m) * n^3/3
- Pi * h/(n-m) * m^3/3
+ h/(n-m) * nd/6 * sqrt(n^2 - d^2)
- h/(n-m) * md/6 * sqrt(m^2 - d^2)
- h/(n-m) * n^3/3 * arccos(d/n)
+ h/(n-m) * m^3/3 * arccos(d/m)
+ h/(n-m) * d^3/6 * log(n + sqrt(n^2 - d^2))
- h/(n-m) * d^3/6 * log(m + sqrt(m^2 - d^2))
+ d * h/(n-m) * n/2 * sqrt(n^2 - d^2)
- d * h/(n-m) * m/2 * sqrt(m^2 - d^2)
- d * h/(n-m) * d^2/2 * log(n + sqrt(n^2 - d^2))
+ d * h/(n-m) * d^2/2 * log(m + sqrt(m^2 - d^2))
You probably didn't expect it to be that complicated!
The value d is the distance from the center to the surface of the
liquid. We can't really use the total depth of the liquid because the
bottom of the tank isn't flat.
This equation only really works when d is between the center than the
top of the smaller circle. If you need to figure it out for other
values of d then please let me know.
Just in case you want to see this work with something you would
recognize the answer to then consider just the first term of the area:
A = Pi R^2
R=n R=n
/ /
V = | A dx = | Pi R^2 dx
/ /
R=m R=m
dx = dR h/(n-m)
R=n
/
V = | Pi R^2 dR h/(n-m)
/
R=m
Then solving the integral we get:
V(R) = Pi * R^3/3 * h/(n-m) where R goes from m to n
Volume = V(n) - V(m)
Volume = Pi * n^3/3 * h/(n-m) - Pi * m^3/3 * h/(n-m)
Volume = Pi * h/3 * (n^3 - m^3)/(n-m)
Volume = Pi * h/3 * (n^2 + nm + m^2)
Which _is_ the volume of a full frustum of a cone.
Anyway, even though it shows that the method works, it isn't the
answer to your question.
So the volume of the liquid in your tank is:
__
+ \
+ \
_ + air \
/ \ \
|---|-----------| ---
| | | d
----+ | |--- --- --- ---
| | liquid | | |
| | | m |
\_/ / | |
+ / --- n
+ / |
+__/ |
---
|----h----|
Volume = Pi * h/(n-m) * (n^3 - m^3)/3
+ h/(n-m) * nd/6 * sqrt(n^2 - d^2)
- h/(n-m) * md/6 * sqrt(m^2 - d^2)
- h/(n-m) * n^3/3 * arccos(d/n)
+ h/(n-m) * m^3/3 * arccos(d/m)
+ h/(n-m) * d^3/6 * log(n + sqrt(n^2 - d^2))
- h/(n-m) * d^3/6 * log(m + sqrt(m^2 - d^2))
+ d * h/(n-m) * n/2 * sqrt(n^2 - d^2)
- d * h/(n-m) * m/2 * sqrt(m^2 - d^2)
- d * h/(n-m) * d^2/2 * log(n + sqrt(n^2 - d^2))
+ d * h/(n-m) * d^2/2 * log(m + sqrt(m^2 - d^2))
- Doctor Jeremiah, The Math Forum
http://mathforum.org/dr.math/